Ionization energy of gaseous $Na$ atoms is $495.5 \ kJ \ mol^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $(h = 6.626 \times 10^{-34} \ J \ s, N_A = 6.022 \times 10^{23} \ mol^{-1})$.
- A$7.50 \times 10^4 \ s^{-1}$
- B$4.76 \times 10^{14} \ s^{-1}$
- C$3.15 \times 10^{15} \ s^{-1}$
- D$1.24 \times 10^{15} \ s^{-1}$
Explore More
Similar Questions
Identify the element which has the highest second ionisation enthalpy.
Which among the following elements has the lowest value of $IE_1$?
Medium
View SolutionExplain why there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.
Medium
View SolutionMatch List-$I$ with List-$II$ with correct code:
List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$) List-$II$ (Element) $A$. $1510$ $1$. $H$ $B$. $495, 6500, 10200$ $2$. $Li$ $C$. $840, 1630, 13100$ $3$. $Be$ $D$. $600, 2050, 3100$ $4$. $B$
| List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$) | List-$II$ (Element) |
|---|---|
| $A$. $1510$ | $1$. $H$ |
| $B$. $495, 6500, 10200$ | $2$. $Li$ |
| $C$. $840, 1630, 13100$ | $3$. $Be$ |
| $D$. $600, 2050, 3100$ | $4$. $B$ |
Difficult
View SolutionThe sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:
Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$ $P$ $2.45$ $8.82$ $Q$ $2.85$ $6.11$
Then,according to the given information,the incorrect statement$(s)$ is/are:
| Element | $IE_1 + IE_2$ $(kJ/mol)$ | $IE_3 + IE_4$ $(kJ/mol)$ |
|---|---|---|
| $P$ | $2.45$ | $8.82$ |
| $Q$ | $2.85$ | $6.11$ |
Then,according to the given information,the incorrect statement$(s)$ is/are:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo